How to find the volume of a tetrahedron formula. Volume of a tetrahedron. Calculating the volume of a tetrahedron if the coordinates of its vertices are known

From the basic formula for the volume of a tetrahedron

Where S is the area of ​​any face, and H– the height lowered by it, a whole series of formulas can be derived that express the volume through various elements of the tetrahedron. Let us present these formulas for the tetrahedron ABCD.

(2) ,

where ∠ ( AD,ABC) – angle between edge AD and the plane of the face ABC;

(3) ,

where ∠ ( ABC,ABD) – angle between faces ABC And ABD;

where | AB,CD| – distance between opposite ribs AB And CD, ∠ (AB,CD) is the angle between these edges.

Formulas (2)–(4) can be used to find the angles between straight lines and planes; formula (4) is especially useful, with which you can find the distance between crossing lines AB And CD.

Formulas (2) and (3) are similar to formula S = (1/2)ab sin C for the area of ​​the triangle. Formula S = rp similar formula

Where r is the radius of the inscribed sphere of the tetrahedron, Σ is its total surface (the sum of the areas of all faces). There is also a beautiful formula connecting the volume of a tetrahedron with the radius R its described sphere ( Crellet formula):

where Δ is the area of ​​a triangle whose sides are numerically equal to the products of opposite edges ( AB× CD, A.C.× BD,AD× B.C.). From formula (2) and the cosine theorem for trihedral angles (see Spherical trigonometry), we can derive a formula similar to Heron’s formula for triangles.

Note. This is part of a lesson with geometry problems (section stereometry, problems about the pyramid). If you need to solve a geometry problem that is not here, write about it in the forum. In tasks, instead of the "square root" symbol, the sqrt() function is used, in which sqrt is the square root symbol, and the radical expression is indicated in brackets.For simple radical expressions, the sign "√" can be used. Regular tetrahedron- This is a regular triangular pyramid in which all faces are equilateral triangles.

In a regular tetrahedron, all dihedral angles at the edges and all trihedral angles at the vertices are equal

A tetrahedron has 4 faces, 4 vertices and 6 edges.

The basic formulas for a regular tetrahedron are given in the table.

Where:
S - Surface area of ​​a regular tetrahedron
V - volume
h - height lowered to the base
r - radius of the circle inscribed in the tetrahedron
R - circumradius
a - edge length

Practical examples

Task.
Find the surface area of ​​a triangular pyramid with each edge equal to √3

Solution.
Since all the edges of a triangular pyramid are equal, it is regular. The surface area of ​​a regular triangular pyramid is S = a 2 √3.
Then
S = 3√3

Answer: 3√3

Task.
All edges of a regular triangular pyramid are equal to 4 cm. Find the volume of the pyramid

Solution.
Since in a regular triangular pyramid the height of the pyramid is projected to the center of the base, which is also the center of the circumscribed circle, then

AO = R = √3 / 3 a
AO = 4√3 / 3

So the height of the pyramid OM can be found from the right triangle AOM

AO 2 + OM 2 = AM 2
OM 2 = AM 2 - AO 2
OM 2 = 4 2 - (4√3 / 3) 2
OM 2 = 16 - 16/3
OM = √(32/3)
OM = 4√2 / √3

We find the volume of the pyramid using the formula V = 1/3 Sh
In this case, we find the area of ​​the base using the formula S = √3/4 a 2

V = 1/3 (√3 / 4 * 16) (4√2 / √3)
V = 16√2/3

Answer: 16√2 / 3 cm

Consider an arbitrary triangle ABC and a point D not lying in the plane of this triangle. Let's connect this point with the vertices of triangle ABC using segments. As a result, we get triangles ADC, CDB, ABD. The surface bounded by four triangles ABC, ADC, CDB and ABD is called a tetrahedron and is designated DABC.
The triangles that make up a tetrahedron are called its faces.
The sides of these triangles are called the edges of the tetrahedron. And their vertices are the vertices of a tetrahedron

The tetrahedron has 4 faces, 6 ribs And 4 peaks.
Two edges that do not have a common vertex are called opposite.
Often, for convenience, one of the faces of a tetrahedron is called basis, and the remaining three faces are side faces.

Thus, a tetrahedron is the simplest polyhedron whose faces are four triangles.

But it is also true that any arbitrary triangular pyramid is a tetrahedron. Then it is also true that a tetrahedron is called a pyramid with a triangle at its base.

Height of tetrahedron called a segment that connects a vertex with a point located on the opposite face and perpendicular to it.
Median of a tetrahedron called a segment that connects a vertex to the point of intersection of the medians of the opposite face.
Bimedian of a tetrahedron called a segment that connects the midpoints of the intersecting edges of a tetrahedron.

Since a tetrahedron is a pyramid with a triangular base, the volume of any tetrahedron can be calculated using the formula

• S– area of ​​any face,
• H– height lowered to this face

Regular tetrahedron - a special type of tetrahedron

A tetrahedron in which all faces are equilateral is called a triangle. correct.
Properties of a regular tetrahedron:

• All edges are equal.
• All plane angles of a regular tetrahedron are 60°
• Since each of its vertices is the vertex of three regular triangles, the sum of the plane angles at each vertex is 180°
• Any vertex of a regular tetrahedron is projected into the orthocenter of the opposite face (at the point of intersection of the altitudes of the triangle).

Let us be given a regular tetrahedron ABCD with edges equal to a. DH is its height.
Let us make additional constructions BM - the height of the triangle ABC and DM - the height of the triangle ACD.
The height of BM is equal to BM and is equal to
Consider the triangle BDM, where DH, which is the height of the tetrahedron, is also the height of this triangle.
The height of the triangle dropped to side MB can be found using the formula

, Where
BM=, DM=, BD=a,
p=1/2 (BM+BD+DM)=
Let's substitute these values ​​into the height formula. We get


Let's take out 1/2a. We get

Let's apply the difference of squares formula

After small transformations we get


The volume of any tetrahedron can be calculated using the formula
,
Where ,

Substituting these values, we get

Thus, the volume formula for a regular tetrahedron is

Where a–tetrahedron edge

Calculating the volume of a tetrahedron if the coordinates of its vertices are known

Let us be given the coordinates of the vertices of the tetrahedron

From the vertex we draw the vectors , , .
To find the coordinates of each of these vectors, subtract the corresponding beginning coordinate from the end coordinate. We get

Definition of tetrahedron

Tetrahedron– the simplest polyhedral body, the faces and base of which are triangles.

Online calculator

A tetrahedron has four faces, each of which is formed by three sides. The tetrahedron has four vertices, with three edges coming out of each.

This body is divided into several types. Below is their classification.

1. Isohedral tetrahedron- all its faces are identical triangles;
2. Orthocentric tetrahedron- all heights drawn from each vertex to the opposite face are equal in length;
3. Rectangular tetrahedron- edges emanating from one vertex form an angle of 90 degrees with each other;
4. Frame;
5. Proportionate;
6. Incentric.

Tetrahedron volume formulas

The volume of a given body can be found in several ways. Let's look at them in more detail.

Through the mixed product of vectors

If a tetrahedron is built on three vectors with coordinates:

A ⃗ = (a x , a y , a z) \vec(a)=(a_x, a_y, a_z)a= (a x​ , a y​ , a z​ )
b ⃗ = (b x , b y , b z) \vec(b)=(b_x, b_y, b_z)b= (b x​ , b y​ , b z​ )
c ⃗ = (c x , c y , c z) \vec(c)=(c_x, c_y, c_z)c= (c x​ , c y​ , c z​ ) ,

then the volume of this tetrahedron is the mixed product of these vectors, that is, the following determinant:

Volume of a tetrahedron through the determinant

V = 1 6 ⋅ ∣ a x a y a z b x b y b z c x c y c z ∣ V=\frac(1)(6)\cdot\begin(vmatrix) a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \\ \end(vmatrix )V=6 1 ​ ⋅ ∣ ∣ ∣ ∣ ∣ ∣ ​ a x​ b x​ c x​ ​ a y​ b y​ c y​ ​ a z​ b z​ c z​ ​ ∣ ∣ ∣ ∣ ∣ ∣ ​

Problem 1

The coordinates of the four vertices of the octahedron are known. A(1, 4, 9) A(1,4,9) A(1, 4, 9), B (8, 7, 3) B(8,7,3) B(8, 7, 3), C (1 , 2 , 3) ​​C (1,2,3) C(1, 2, 3), D(7, 12, 1) D(7,12,1) D(7, 1 2, 1). Find its volume.

Solution

A(1, 4, 9) A(1,4,9) A(1, 4, 9)
B (8, 7, 3) B(8,7,3) B(8, 7, 3)
C (1 , 2 , 3) ​​C (1,2,3) C(1, 2, 3)
D(7, 12, 1) D(7,12,1) D(7, 1 2, 1)

The first step is to determine the coordinates of the vectors on which this body is built.
To do this, you need to find each vector coordinate by subtracting the corresponding coordinates of the two points. For example, the vector coordinates A B → \overrightarrow(AB) A B, that is, a vector directed from the point A A A to the point B B B, these are the differences between the corresponding coordinates of the points B B B And A A A:

A B → = (8 − 1 , 7 − 4 , 3 − 9) = (7 , 3 , − 6) \overrightarrow(AB)=(8-1, 7-4, 3-9)=(7, 3, -6)A B= (8 − 1 , 7 − 4 , 3 − 9 ) = (7 , 3 , − 6 )

A C → = (1 − 1 , 2 − 4 , 3 − 9) = (0 , − 2 , − 6) \overrightarrow(AC)=(1-1, 2-4, 3-9)=(0, - 2, -6)A C= (1 − 1 , 2 − 4 , 3 − 9 ) = (0 , − 2 , − 6 )
A D → = (7 − 1 , 12 − 4 , 1 − 9) = (6 , 8 , − 8) \overrightarrow(AD)=(7-1, 12-4, 1-9)=(6, 8, -8)A D= (7 − 1 , 1 2 − 4 , 1 − 9 ) = (6 , 8 , − 8 )

Now let’s find the mixed product of these vectors; to do this, we’ll compose a third-order determinant, while accepting that A B → = a ⃗ \overrightarrow(AB)=\vec(a)A B= a, A C → = b ⃗ \overrightarrow(AC)=\vec(b)A C= b, A D → = c ⃗ \overrightarrow(AD)=\vec(c)A D= c.

a x a y a z b x b y b z c x c y c z (− 6) ⋅ (− 2) ⋅ 6 − 7 ⋅ (− 6) ⋅ 8 − 3 ⋅ 0 ⋅ (− 8) = 112 − 108 − 0 − 72 + 336 + 0 = 268 \begin(vmatrix) a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \\ \end(vmatrix)= \begin(vmatrix) 7 & 3 & -6 \\ 0 & -2 & -6 \\ 6 & 8 & -8 \\ \end(vmatrix)=7\cdot(-2)\cdot(-8) + 3\cdot(-6)\cdot6 + (-6)\cdot0\cdot8 - (-6)\cdot (-2)\cdot6 - 7\cdot(-6)\cdot8 - 3\cdot0\cdot(-8) = 112 - 108 - 0 - 72 + 336 + 0 = 268∣ ∣ ∣ ∣ ∣ ∣ ​ a x​ b x​ cx​ ​ ay​ by​ cy​ ​ az​ bz​ cz​ ​ ∣ ∣ ∣ ∣ ∣ ∣ ​ = ∣ ∣ ∣ ∣ ∣ ∣ ​ 7 0 6 ​ 3 − 2 8 ​ − 6 − 6 − 8 ​ ∣ ∣ ∣ ∣ ∣ ∣ ​ = 7 ⋅ (− 2 ) ⋅ (− 8 ) + 3 ⋅ (− 6 ) ⋅ 6 + (− 6 ) ⋅ 0 ⋅ 8 − (− 6 ) ⋅ (− 2 ) ⋅ 6 − 7 ⋅ (− 6 ) ⋅ 8 − 3 ⋅ 0 ⋅ (− 8 ) = 1 1 2 − 1 0 8 − 0 − 7 2 + 3 3 6 + 0 = 2 6 8

That is, the volume of the tetrahedron is equal to:

$V=61\cdot |$

Answer

$44.8\text{cm3 .}$

Formula for the volume of an isohedral tetrahedron along its side

This formula is valid only for calculating the volume of an isohedral tetrahedron, that is, a tetrahedron in which all faces are identical regular triangles.

Volume of an isohedral tetrahedron

$V=12\sqrt{2\cdot a^3}$

a aa- length of the edge of the tetrahedron.

Problem 2

Determine the volume of a tetrahedron given its side equal to $11\text{cm.}$

Solution

$a=11$

Let's substitute a aa into the formula for the volume of a tetrahedron:

$V=12\sqrt{2\cdot a^{3=12\sqrt{2\cdot 11^{3\approx 156.8\text{cm3}}}}}$

Answer

$156.8\text{cm3 .}$